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Условие Отговори Решение

USAPhO 2003, Четвъртфинал — Задача 4

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Всички материали от USAPhO 2003

AAPT AIP United States Physics Team 2003 — Preliminary Exam, Open Response Questions · 25 т.

Указания

4 Questions, 60 minutes

INSTRUCTIONS

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example,

Doe, Jamie

Q1 – 1/3
A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. You may not use any tables, books, or collections of formulas.
Each of the four questions is worth 25 points. The questions are not necessarily of the same difficulty.
Good luck!

Possibly Useful Information

Gravitational field at the Earth’s surface	$g = 9.8\ \mathrm{N/kg}$
Newton’s gravitational constant	$G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$

Moment of Inertia about Center of Mass – Uniform Object

(will not be provided on the second screening exam)

Disk	$\tfrac{1}{2} MR^2$
Sphere	$\tfrac{2}{5} MR^2$
Rod	$\tfrac{1}{12} ML^2$

Solutions to Problems

Unless explicitly told to use a specific method, any correct solution should be awarded equivalent points. Suggested partial-credit points are presented in square brackets at the right margin. You may further break down the listed points into one point increments. If it is clear they have done an intermediate step, they should get credit for it even if they have not presented it. For example, in 2.a., if a student wrote down $2v_o t - \frac{1}{2}gt^2 = H - v_o t - \frac{1}{2}gt^2$ , they should get 8 points credit. Students should not be penalized in a subsequent part for using the wrong answer to a previous part. (No double jeopardy.)

Условие

A circle of radius R labelled M resting on a horizontal belt line; below the belt a black arrow pointing right labelled v.

A solid, uniform cylinder with mass $M$ and radius $R$ is initially at rest. It is gently placed onto a conveyor belt moving at constant velocity $v$ as shown above. The cylinder’s axis is perpendicular to the belt’s velocity. The coefficients of static and kinetic friction are nonzero constants. (Ignore other energy losses such as rolling resistance.)

a) What is the final speed of the cylinder’s center of mass? [15 т.]

b) If a cylinder with twice the radius and four times the mass had been used in place of the original cylinder, what would have been its final speed? [5 т.]

c) If a ring with the same mass and radius had been used in place of the cylinder, what would have been its final speed? Assume the ring has negligible thickness. [5 т.]

Отговори

Решение

4. a. When the cylinder is first placed on the conveyor belt, it is slipping and will experience a force of kinetic friction $f$ . This causes both an acceleration of the center of mass and an angular acceleration about the center of mass. The cylinder will continue to accelerate until it has reached a velocity at which it rolls without slipping. Then the net force is zero. [2]

When it rolls without slipping,

v = v_{cm} + v_{rel} \qquad (4-1)

[2]

where $v_{cm}$ is the final velocity of the center of mass and

v_{rel} = \omega R

[1]

is the velocity of a point on the rim relative to the center of mass.

(There are at least three methods that can be used to find the relation between $v_{cm}$ and $\omega R$ . These are detailed below. Students may be awarded points from only one method.)

Method I – Using the acceleration to find the velocities.

(It is not necessary to set $f = \mu Mg$ . In all of the following answers $\mu g$ and $f/M$ may be used interchangeably.) Finding the translational acceleration of the center of mass

\sum F = Ma_{cm}

[1]

f = \mu Mg = Ma_{cm}

a_{cm} = \mu g.

[1]

Since it starts from rest, after time $t$ , the velocity of the center of mass is

v_{cm} = \mu gt.

[1]

Finding the angular acceleration about the center of mass

\sum \tau = I\alpha.

[1]

The torque is

\tau = Rf = \mu RMg.

[1]

The moment of inertia of a solid uniform cylinder is

I = \tfrac{1}{2}MR^2

[1]

\alpha = \frac{\tau}{I} = \frac{fR}{\frac{1}{2}MR^2} = \frac{2\mu Mg}{MR} = \frac{2\mu g}{R}.

[1]

After time $t$ , the angular velocity is

\omega = \alpha t = \frac{2\mu gt}{R}.

[1]

The velocity of a point on the rim relative to the center of mass is

\omega R = 2\mu gt = 2v_{cm}.

[1]

Method II – Using impulse to find the velocities.

The magnitude of the impulse is given by

\Delta p_{cm} = F_{net}t

[1]

where $t$ is the time for which the force acts. The change in momentum is

\Delta p_{cm} = Mv_{cm} - 0.

[1]

Mv_{cm} = ft. \qquad (4-2)

[1]

The magnitude of the angular impulse is

\Delta L = \tau t.

[1]

The change in angular momentum is

\Delta L = I\omega - 0.

[1]

The magnitude of the torque is

\tau = fR.

[1]

Combining

I\omega = fRt. \qquad (4-3)

[1]

For a cylinder the moment of inertia is

I = \tfrac{1}{2}MR^2. \qquad (4-4)

[1]

Substituting Eq. (4-2) and (4-4) into (4-3)

\tfrac{1}{2}MR^2\omega = Mv_{cm}R.

[1]

Solving for $\omega R$

\omega R = 2v_{cm}.

Method III. Use conservation of angular momentum.

This is easiest to do in the lab frame. Consider an axis that is at rest in the lab frame and lies along the initial line of contact between the cylinder and the belt. The line of action of the friction force passes through this axis, so friction provides no torque. The normal force and the weight force provide equal and opposite torques about any axis. The net torque about this axis is zero.

\sum \tau = 0.

[2]

The total angular momentum equals a constant $C$

\sum L = C.

[1]

The angular momentum has two components, the angular momentum due to the rotation of the cylinder and the angular momentum due to the translation of the center of mass.

\sum L = L_{rot} + L_{trans}.

These are in equal and opposite senses

\sum L = I\omega - MRv_{cm} = C.

[2]

Since $\omega = 0$ and $v_{cm} = 0$ initially, $C = 0.$ [1]

I\omega = MRv_{cm}

[1]

For a cylinder the moment of inertia is

I = \tfrac{1}{2}MR^2.

[1]

Combining

\tfrac{1}{2}MR^2\omega = Mv_{cm}R.

[1]

Solving for $\omega R$

\omega R = 2v_{cm}.

Then for all methods:

Substituting into Eq. (4.1)

v = v_{cm} + \omega R.

v = v_{cm} + 2v_{cm}.

v_{cm} = \tfrac{1}{3}v.

[1]

b. The velocity does not depend on $M$ or $R$ , therefore

v_{cm} = \tfrac{1}{3}v.

[5]

c. If a ring of infinitesimal thickness were used, the moment of inertia and angular velocity would change. Now

I = MR^2,

[2]

v_{rel} = \omega R = \alpha tR = \frac{\tau}{I}tR = \frac{\mu RMg}{MR^2}tR = \mu gt

[1]

v = v_{cm} + v_{rel} = \mu gt + \mu gt

v_{cm} = \tfrac{1}{2}v.

[2]

Оригинал в Архива: Quarterfinal 2003.pdf · официални решения: Quarterfinal 2003 Solutions.pdf